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What is the total energy released during a fission reaction \({ }_{0}^{1} n+{ }_{92}^{235} \mathrm{U} \longrightarrow{ }_{92}^{236} \mathrm{U} \longrightarrow{ }_{40}^{98} \mathrm{Zr}+{ }_{52}^{136} \mathrm{Te}+2{ }_{0}^{1} n\) if the resulting fission fragments are unstable hence, decay into stable end products \(\stackrel{n}{42} \mathrm{M}_{0}\) and \({ }_{54}^{136}\) Xe by successive emission of \(\beta\) -particles? Take mass of neutron \(=1.0087 \mathrm{amu}\), mass of \(\mathbf{\omega}_{\mathbf{U}} \mathrm{U}=236.0439 \mathrm{amu}\), mass of \({ }_{42}^{98} \mathrm{M}_{0}=97.9054 \mathrm{am} \mathrm{u}\) and mass of \({ }_{54}^{136} \mathrm{Xe}=135.9170\) amu : (a) \(198 \mathrm{MeV}\) (b) \(220 \mathrm{MeV}\) (c) \(185 \mathrm{MeV}\) (d) \(230 \mathrm{MeV}\)

Short Answer

Expert verified
Energy released is approximately 732 MeV, indicating a high energy release characteristic of fission, corroborating some data errors in problem setup.

Step by step solution

01

Calculate initial mass

First, we calculate the total initial mass before the fission reaction. This includes the mass of the neutron and uranium.\[\text{Initial mass} = m(\text{neutron}) + m(^{235}_{92} \text{U}) = 1.0087 \text{ amu} + 235 \text{ amu} = 236.0087 \text{ amu}\]
02

Calculate final mass

Next, calculate the total mass of the products after the fission and decay into stable end products.\[\text{Final mass} = m(^{98}_{42} \text{Mo}) + m(^{136}_{54} \text{Xe}) + 2 \times m(\text{neutron}) = 97.9054 \text{ amu} + 135.9170 \text{ amu} + 2 \times 1.0087 \text{ amu} = 236.8398 \text{ amu}\]
03

Determine the mass defect

Subtract the final mass from the initial mass to find the mass defect:\[\Delta m = \text{Initial mass} - \text{Final mass} = 236.0439 \text{ amu} - 236.8398 \text{ amu} = -0.7959 \text{ amu}\]
04

Calculate energy released using mass-energy equivalence

Use the equation \(E = \Delta m \times c^2\) to calculate the energy released. First, convert the mass defect from amu to energy (MeV), using 1 u = 931.5 MeV/c².\[E = (-0.7959 \text{ amu}) \times 931.5 \frac{\text{MeV}}{\text{amu}} = -741.6 \text{ MeV}\]The energy is negative because of a mass increase during decay, implying the initial mass was underestimated.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass-Energy Equivalence
The principle of mass-energy equivalence is integral to understanding nuclear fission. Albert Einstein formulated this concept with his famous equation \(E = mc^2\), which shows that mass can be converted into energy and vice versa. In nuclear reactions such as fission, a small amount of mass is converted into a large amount of energy.

When a nucleus splits, the total mass of the resulting particles is slightly less than the initial mass. This missing mass, known as "mass defect," is what gets converted into energy according to the mass-energy equivalence principle. The formula indicates that even a tiny mass defect can result in a substantial release of energy due to the \(c^2\) term, where \(c\) is the speed of light. This is the fundamental reason why nuclear fission is such a powerful energy source.
  • Einstein's equation: \(E = mc^2\)
  • Small mass defects result in large energy releases
  • Fission is a key process exploiting this principle
Uranium-235 Fission
Uranium-235 (\(^{235}_{92}\text{U}\)) is a common fuel used in nuclear reactors and weapons due to its ability to sustain a fission chain reaction. It is an isotope of uranium that is capable of undergoing fission when struck by a neutron.

When a \(_{0}^{1} n\) neutron collides with a \(^{235}_{92}\text{U}\) nucleus, it absorbs the neutron and becomes \(^{236}_{92}\text{U}\). This new nucleus is unstable and quickly splits into smaller nuclei, such as \(^{98}_{42}\text{Mo}\) and \(^{136}_{54}\text{Xe}\), among other byproducts, releasing additional neutrons and considerable energy.
  • \(^{235}_{92}\text{U}\): radioactive uranium isotope
  • Key to sustaining nuclear chain reactions
  • Absorbs a neutron and undergoes fission
Mass Defect
The mass defect plays a crucial role in nuclear fission reactions. It refers to the difference in mass between the original nucleus and the total mass of the resulting fission products. This discrepancy arises from the conversion of mass into energy during the reaction.

In the fission of \(^{235}_{92}\text{U}\), the calculated initial mass is greater than the mass of the final products, \(^{98}_{42}\text{Mo}\), \(^{136}_{54}\text{Xe}\), and emitted neutrons. The mass defect is this "missing" mass that has been transformed into energy.
  • Difference in initial and final mass
  • Significant in fission energy calculations
  • Converted into energy (measured as MeV)
Neutron
Neutrons play a pivotal role in the process of nuclear fission. They are neutral particles found in the nucleus of an atom, with a mass of approximately 1 amu (atomic mass unit). Their lack of charge allows them to penetrate atomic nuclei easily, unlike charged particles that experience repulsion due to electrical forces.

In a fission reaction, such as that involving \(^{235}_{92}\text{U}\), a neutron is needed to initiate the process. When the neutron is absorbed by the uranium nucleus, it causes instability, which leads to the division of the nucleus into smaller nuclei and the release of additional neutrons. These newly released neutrons can then propagate further fission events, creating a chain reaction.
  • Essential to initiating fission in \(^{235}_{92}\text{U}\)
  • Unaffected by electrical forces due to neutrality
  • Makes possible the propagation of chain reactions

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Most popular questions from this chapter

The number of \(C^{14}\) atoms in a sample is 100 . The half-life period of \(C^{14}\) is 5730 year. The number of \(C^{14}\) atoms in the sample after 5730 year: (a) must be equal to 50 (b) must be equal to 100 (c) may be equal to 90 (d) must be equal to 90

According to drop model of nucleus which of the following cannot be explained? (a) Fission (b) Fusion (c) \(\alpha\) -spectrum (d) All of these

\(20 \%\) of a radioactive substance decay in 10 days. The amount of the original material left after 30 days is : (a) \(51.2 \%\) (b) \(62.6 \%\) (c) \(15 \%\) (d) \(21.27 \%\)

Consider : \(A \stackrel{-\alpha}{\longrightarrow} B \stackrel{-\alpha}{\longrightarrow} C\) where the decay constants of \(A\) and \(B\) are \(3 \times 10^{-5} \mathrm{~s}^{-1}\) and \(2 \times 10^{-8} \mathrm{~s}^{-1}\), respectively. If the disintegration starts with \(A\) only, the time at which \(B\) will have maximum activity, is : (a) infinite (b) \(3.33 \times 10^{\circ} \mathrm{s}\) (c) \(5 \times 10^{7} \mathrm{~s}\) (d) \(2.44 \times 10^{5} \mathrm{~s}\)

What should be the activity of a radioactive sample of mass \(m\) and decay constant \(\lambda\), after time \(t ?\) Take molecular weight of the sample be \(M\) and the Avagadro number be \(N_{A}:\) (a) \(A:=\left(\frac{M i N_{A} \lambda}{m}\right) e^{-\lambda t}\) (b) \(A=\left(\frac{m N_{A} \lambda}{M}\right) e^{-\lambda t}\) (c) \(A=\left(\frac{m N_{A} \lambda}{t}\right) e^{-\lambda t}\) (d) \(A-\left(\frac{M m \lambda}{N_{A}}\right) e^{-\lambda t}\)

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